Ages – father as a linear combination of children: The father of two children has an age equal to twice the elder child’s age plus four times the younger child’s age. The geometric mean (GM) of the children’s ages is 4√3, and their harmonic mean (HM) is 6. Find the father’s age in years.

Introduction / Context:
This is a number-theory/algebra blend using classical means (geometric and harmonic) to deduce two unknown ages. Once the children’s ages are identified, the father’s age is obtained from a linear combination: 2 × (elder) + 4 × (younger).

Given Data / Assumptions:

• Let children’s ages be x (elder) and y (younger), x ≥ y.
• GM = √(xy) = 4√3 ⇒ xy = 48.
• HM = 2xy / (x + y) = 6 ⇒ x + y = 2xy / 6 = 16.
• Father’s age F = 2x + 4y.

Concept / Approach:
Use Vieta-like relationships from the quadratic with sum and product: t^2 − (x + y)t + xy = 0 ⇒ t^2 − 16t + 48 = 0. Solve to get the two ages, then compute F = 2x + 4y.

Step-by-Step Solution:
x + y = 16, xy = 48.t^2 − 16t + 48 = 0 ⇒ Discriminant = 256 − 192 = 64.t = (16 ± 8) / 2 ⇒ t ∈ {12, 4} ⇒ (elder, younger) = (12, 4).F = 2*12 + 4*4 = 24 + 16 = 40 yr.

Verification / Alternative check:
GM = √(12*4) = √48 = 4√3; HM = 2*12*4/(12+4) = 96/16 = 6 — both match, confirming ages 12 and 4 and hence F = 40 yr.

Why Other Options Are Wrong:
32, 48, 56, 44 yr do not equal 2*12 + 4*4 and do not satisfy the derived constraints simultaneously.

Common Pitfalls:
Mixing up which mean formula to use; computing GM or HM incorrectly; reversing coefficients in F.

Final Answer:
40 yr